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+---
+author: 'Yaël Dillies, Paul Lezeau'
+category: 'Metaprogramming'
+date: 2025-04-16 12:00:00 UTC+01:00
+description: 'How to write a simproc in practice'
+has_math: true
+link: ''
+slug: simprocs-tutorial
+tags: 'simp, simproc, meta'
+title: 'Simprocs, the process made simple'
+---
+
+This is the final post in our simproce series. In our first two posts, we gave an informal introduction to the concept of a *simproc*, and a brief overview of the inner workings of the `simp` tactic. The aim of this final post is to use build on this by demonstrating how Lean users can write their own simprocs.
+
+> As for the previous post, we will assume that the reader has some exposure to metaprogramming in Lean.
+> In addition, some familiarity with the `Qq` library will be helpful, but not necessary for most of this post.
+
+Then we explain the syntax and general structure of a simproc.
+Finally, we walk through an explicit example of a simproc for a simple custom function.
+
+# The simproc syntax
+
+Let's see how to declare a simproc.
+
+The basic syntax for declaring a simproc is
+```lean
+simproc_decl mySimproc (theExprToMatch _ _) := fun e ↦ do
+  write_simproc_here
+```
+> See the next section for how to actually replace `write_simproc_here` by the correct meta code.
+
+To add `mySimproc` to the standard simp set, replace `simproc_decl` by `simproc`:
+```lean
+simproc mySimproc (theExprToMatch _ _) := fun e ↦ do
+  write_simproc_here
+```
+
+When calling a simproc in `simp` (if it is not in the standard simp set), one can specify that this is a preprocedure by adding `↓` in front of the simproc identifier: `simp [↓mySimproc]` (note that this also works when passing lemmas to `simp`!)
+
+To add `mySimproc` to the standard simp set as a preprocedure (recall that postprocedure is the default), do
+```lean
+simproc ↓ mySimproc (theExprToMatch _ _) := fun e ↦ do
+  write_simproc_here
+```
+Note that being a pre/postprocedure is a property of simprocs *in a simp set*, not of bare simprocs.
+Therefore, there is no corresponding `simproc_decl ↓` syntax.
+
+# Simproc walkthrough
+
+Let's write a simproc for a simple recursive function.
+We choose a custom function `revRange`, which to a natural number `n` returns the list of the first `n` natural numbers in decreasing order:
+
+```lean
+def revRange : Nat → List Nat
+  | 0     => []
+  | n + 1 => n :: revRange n
+
+#eval revRange 5 -- [4, 3, 2, 1, 0]
+```
+
+Our goal will be to make `simp` evaluate `revRange` when its input is an explicit numeral, eg
+```lean
+example : revRange 0 = [] := by simp [???]
+example : revRange 2 = [1, 0] := by simp [???]
+example : revRange 5 = [4, 3, 2, 1, 0] := by simp [???]
+```
+
+Note two features of `revRange` that one should *not* expect from all functions that one might want to evaluate on explicit inputs:
+* It is **recursive**: One can compute `revRange n` by recursion on `n`.
+  Even more precisely, `revRange n` represents its own partial computation.
+* `revRange` is definitionally equal to what we want to unfold it to.
+  This has two consequences:
+  * The two examples in the code snippet above can be proved by `rfl`, but of course doing so defeats the point of this blogpost.
+  * We could actually write a *dsimproc* for `revRange`, which is to `dsimp` what a simproc is to `simp`.
+    Implementation-wise, the main difference is that a dsimproc requires the new simplified expression to be definitionally equal to the previous one.
+
+Let's now present three approaches to evaluating `revRange` on numerals:
+* The baseline **simproc-less approach** which only uses lemmas and no simproc.
+* The **dsimproc approach**, where we (possibly recursively) construct in the meta world the evaluated expression, but leave the proof to be `rfl` (here the "d" stands for "definitional equality").
+* The **simproc approach**, where we (possibly recursively) construct the evaluated expression and the proof simultaneously.
+
+## The simproc-less approach
+
+Before writing a simproc, let us first see how one could approach the computation of `revRange` using only lemmas.
+
+`revRange` is a recursive function.
+Therefore it can be evaluated on numerals simply by writing out the recurrence relations we wish to reduce along:
+```lean
+lemma revRange_zero : revRange 0 = [] := rfl
+lemma revRange_succ (n : Nat) : revRange (n + 1) = n :: revRange n := rfl
+```
+
+Then we can complete our code snippet like so:
+```lean
+example : revRange 0 = [] := by simp [revRange_zero, revRange_succ]
+example : revRange 2 = [1, 0] := by simp [revRange_zero, revRange_succ]
+example : revRange 5 = [4, 3, 2, 1, 0] := by simp [revRange_zero, revRange_succ]
+```
+
+Note: Since `revRange` is defined by recursion, `simp [revRange]` would also be a valid proof here.
+But we are trying not to rely on the definition of `revRange`.
+
+**Pros**:
+* Doesn't require writing any meta code.
+* Doesn't require the recursion relations to be definitional (although they are in the case of `revRange`).
+
+**Cons**:
+* Requires adding two lemmas to your simp call instead of one (assuming we do not want these lemmas in the default simp set).
+* Simplifying `revRange n` for a big input numeral `n` might involve a lot of simplification steps.
+  In this specific case, the number of simplification steps is linear in `n`.
+  Simplification steps matter because each of them increases the size of the proof term.
+* `revRange n` could find itself (partially) evaluated even if `n` isn't a numeral.
+  Eg `simp [revRange_zero, revRange_succ]` on `⊢ revRange (n + 3) = revRange (3 + n)` will result in `⊢ n + 2 :: n + 1 :: n :: revRange n = revRange (3 + n)`.
+  This is in general highly undesirable.
+
+## The definitional approach
+
+In cases where the evaluation is definitionally equal to the original expression, one may write a dsimproc instead of a simproc.
+The syntax to declare a dsimproc is rather to simprocs, with a small difference: we now need to return a [`DStep`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Meta.Simp.DStep#doc) instead of a `Step`; in practice this amounts to providing the expression our program has produced without providing the proof (indeed, this is just `rfl`!)
+
+<span style="color:red">**TODO**: We were explaining `DStep` before, but now it comes after.</span>
+
+To compute `revRange` using the dsimproc approach, we can do the following:
+```lean
+dsimproc_decl revRangeCompute (revRange _) := fun e => do
+  -- Extract the natural number from the expression
+  let_expr revRange m ← e | return .continue
+  -- Recover the natural number as a term of type `Nat`
+  let some n := m.nat? | return .continue
+  let l := revRange n
+  -- Convert the list to an `Expr`
+  return .visit <| Lean.toExpr l
+```
+
+For a bit more on dsimprocs, see the extras below.
+
+**Pros**:
+* Requires writing a single simproc.
+* Assuming the type of the expression to be evaluated implements `ToExpr`, there is no need to reevaluate the expression manually in the meta world.
+
+**Cons**:
+* The function needs to be computable to be evaluated automatically in the meta world.
+* The produced `rfl` proof could be heavy.
+* Only works when the evaluation and conversion back to an expression is definitionally equal to the original expression.
+
+## The propositional approach
+
+A more general approach would be to manually construct the proof term we need to provide.
+In our case, we can do this in a recursive manner.
+```
+open Qq
+
+private theorem revRange_succ_eq_of_revRange_eq {n : ℕ} {l : List ℕ}
+    (hl : revRange n = l) : revRange (n+1) = n :: l := by
+  induction n with
+  | zero => aesop
+  | succ n h => rw [←hl]; rfl
+
+open Qq in
+simproc_decl revRangeComputeProp (revRange _) := fun e => do
+  let_expr revRange m ← e | return .continue
+  let some n ← Nat.fromExpr? m | return .continue
+  let rec go (n : ℕ) : (l : Q(List ℕ)) × Q(revRange $n = $l) :=
+    match n with
+    | 0 => ⟨q(([] : List ℕ)), q(revRange_zero)⟩
+    | n + 1 =>
+      let ⟨l, pf⟩ := go n
+      ⟨q($n :: $l), q(revRange_succ_eq_of_revRange_eq $pf)⟩
+  let ⟨l, pf⟩ := go n
+  return .visit { expr := l, proof? := pf }
+```
+
+**Pros**:
+* Works regardless of definitional equalities.
+  See [`Nat.reduceDvd`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Nat.reduceDvd#doc) introduced in [the previous blog post](https://leanprover-community.github.io/blog/posts/fantastic-simprocs/) for another compelling example:
+  `a ∣ b` is *not* defined as `a % b = 0`, yet the `Nat.reduceDvd` simproc can decide `a ∣ b` by computing `a % b = 0`.
+
+**Cons**:
+* Might involve a fair bit of meta code, a lot of which could *feel* like evaluating the function.
+* Simplifying `revRange n` for a big input numeral `n` might produce a large proof term.
+  In this specific case, the size of the produced proof term will be linear in `n`.
+
+# Extras
+
+## How to discharge subgoals
+
+Often, when applying a theorem, we may need to provide additional proof terms for the hypotheses of the result. One useful feature of
+`simprocs` is that we can also call the discharger tactic provided to simp. Which discharger was provided by the user is part of the state stored by the `SimpM` monad, and can be access by the user via [`Methods`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Meta.Simp.Methods#doc) (roughly speaking, the part of the state that encodes which methods `simp` can use to simplify an expression). `Methods` implements a function `discharge? : Expr → Option Expr` such that `discharge? goal` is equal to `some pf` if the discharger found a proof `pf` of `goal`, and none otherwise. Finally, to access the current "state" of `Methods`, one can use [`getMethods`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Meta.Simp.getMethods#doc).
+
+In the following example, we implement a simproc for [`Nat.factorization`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Nat.factorization#doc) that simplifies expressions of the form `(a * b).factorization ` to `a.factorization + b.factorization` whenever a proof that `a` and `b` are both non-zero can be found by the discharger.
+
+```lean
+import Mathlib
+
+open Qq Lean.Meta.Simp
+
+simproc_decl factorizationMul (Nat.factorization (_ * _)) := .ofQ fun u α e => do
+  match u, α, e with
+  | 1, ~q(ℕ →₀ ℕ), ~q(Nat.factorization ($a * $b)) =>
+    -- Try to discharge the goal `a ≠ 0`
+    let some ha ← ((← getMethods).discharge? q($a ≠ 0)) | return .continue
+    --Convert the resulting proof to a `Qq` expression for convenience (see #23510)
+    let ⟨0, ~q($a ≠ 0), ~q($ha)⟩ ← inferTypeQ ha | return .continue
+    -- Try to discharge the goal `b ≠ 0`
+    let some hb ← ((← getMethods).discharge? q($b ≠ 0)) | return .continue
+    --Convert the resulting proof to a `Qq` expression for convenience
+    let ⟨0, ~q($b ≠ 0), ~q($hb)⟩ ← inferTypeQ hb | return .continue
+    let e' := q((Nat.factorization $a) + (Nat.factorization $b))
+    let pf := q(Nat.factorization_mul $ha $hb)
+    return .visit { expr := e', proof? := pf }
+  | _, _, _ => return .continue
+
+set_option trace.Meta.Tactic.simp true in
+example : Nat.factorization (2 * 3) = fun₀ | 2 => 1 | 3 => 1 := by
+  simp (disch := decide) only [factorizationMul]
+  guard_target = Nat.factorization 2 + Nat.factorization 3 = fun₀ | 2 => 1 | 3 => 1
+  sorry
+```
+
+## How to match on numerals
+
+Often when writing a simproc to perform a computation, it can be useful to extract quantities from the expression we are manipulating. The easiest case is perhaps that of `Nat` litterals. Given a numeral by `e : Expr`, there are various ways of recovering the corresponding term of type `Nat`:
+- [`Lean.Expr.rawNatLit?`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Expr.rawNatLit?#doc).
+- [`Lean.Expr.natLit!`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Expr.rawNatLit!#doc)
+- [`Lean.Expr.nat?`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Lean.Expr.nat?#doc)
+- [`Nat.fromExpr?`](https://leanprover-community.github.io/mathlib4_docs/find/?pattern=Nat.fromExpr?#doc)
+
+<span style="color:red">**TODO(Paul)**: Let's explain the differences, and show some examples of where behaviour differs. </span>
+
+## How to handle a non-recursive definition
+
+Write a type of partial computations that is recursive.
+
+<span style="color:red">**TODO(Paul)**</span>
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